Take group $13$ oxides as an example. The elements in this group are $\ce{B,Al,Ga,In,Tl}$.

Boron oxide $\ce{B2O3}$ is acidic and $\ce{Al2O3}$ and $\ce{GaO3}$ are amphoteric while $\ce{In2O3}$ and $\ce{Tl2O3}$ are basic in nature.

What is the reason for this?

  • $\begingroup$ Metallic character increases down the group. $\endgroup$ May 4, 2017 at 15:51
  • $\begingroup$ But are they really basic? I mean the last two. Aren't they amphoteric? I think they are less acidic because the electronegativity gets lower as you get lower down the group. $\endgroup$ May 4, 2017 at 17:24

1 Answer 1


As suggested in comments, major factors play a part

  1. Atomic size

As the atomic size of the element increases, its electron affinity decreases consequently. It turns out that when moving vertically in the periodic table, the size of the atom outdoes its electronegativity with regard to basicity. The atomic radius of gallium, for example is approximately twice that of boron, so in a gallium ion, the charge is spread out over a significantly larger volume.

Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.

  1. Electron affinity

The ability of an atom “to accept an electron” decreases down the group therefore increases the tendency of the elements to be good Lewis bases (recall a Lewis base is an electron-pair donor). Thus the decrease in electron affinity down the group means the elements readily donate an electron rather than accept them.

  1. Electronegativity

The more electronegative an atom, the better it is able to bear a negative charge. Down the group, the electronegativity of the elements decrease down the group. B is very electronegative and doesn’t readily release electrons but instead holds on to it, additionally is able to stabilise a negative charge, and consequently has good acidic properties unlike proceeding elements down the group.

  • 1
    $\begingroup$ Can you tell how polarizability and basicity are related? $\endgroup$
    – Ka Sikh
    May 7, 2017 at 15:33

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